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Length of job tenure. Researchers at the Terry College ofBusiness at the University of Georgia sampled 344 business students and asked them this question: 鈥淥ver the course of your lifetime, what is the maximum number of years you expect to work for any one employer?鈥 The sample resulted in x= 19.1 years. Assume that the sample of students was randomly selected from the 6,000 undergraduate students atthe Terry College and that = 6 years.

  1. Describe the sampling distribution of X.
  2. If the mean for the 6,000 undergraduate students is= 18.5 years, findPx>19.1.
  3. If the mean for the 6,000 undergraduate students is= 19.5 years, findPx>19.1.
  4. If,P(x>19.1)=0.5 what is?
  5. If,Px>19.1=0.2 isgreater than or less than 19.1years? Explain.

Short Answer

Expert verified
  1. The sampling distribution ofX isx344
  2. If the mean for the 6,000 undergraduate students is = 18.5 years, PX>19.10.0318.
  3. If the mean for the 6,000 undergraduate students is = 19.5 years, find PX>19.10.8919.
  4. IfPX>19.1=0.5,=19.1
  5. If PX>19.1=0.2, is less than 19.1.

Step by step solution

01

Given information

In Terry College of Business at the University of Georgia, some researchers took a sample of 344 students from a population of 6000 undergraduate students randomly. Then they questioned the sampled students that what is the number of maximum years they expect to work for any one employer.

The survey resulted that x=19.1 years. The standard deviation =6years.

02

describe the sampling distribution of the sample mean

The sampling distribution ofx is described as,

Meanx=

Standard deviationx=n=63440.323

z-scorez=xxx=x344

03

Calculate the probability Px¯>19.1 when  μ=18.5 years

We have the mean of the 6000 undergraduate students, that is =18.5.

Here, n=344. So, n is greater than 30. We can say that it is a relatively large sample size.

Therefore, by the Central Limit Theorem we can say that X~N18.5,0.323.

Thus,

PX>19.1=normalcdf19.1,999,18.5,0.3230.0318

04

Calculate the probability Px¯>19.1 when μ =19.5 years

We have the mean of the 6000 undergraduate students, that is =19.5.

Here, n=344. So, n is greater than 30. We can say that it is a relatively large sample size.

Therefore, by the Central Limit Theorem we can say that X~N19.5,0.323.

Thus,

PX>19.1=normalcdf19.1,999,19.5,0.3230.8919

05

Calculate the mean μ when Px¯>19.1=0.5

There is given that PX>19.1=0.5. So, we can conclude that 50% of the values ofX are below 19.1 and 50% are above 19.1.

Therefore, the distribution ofX is symmetric for this case. Thus, the expected value ofX must be 19.1.

Therefore, X=19.1.

Referring to a. part, we know that X=. So, the value of =19.1.

06

Determine the mean

We are considered that PX>19.1=0.2. So, we can conclude that 20% values ofX are above 19.1 and 80% of values are below 19.1.

Therefore, the distribution is not symmetric for this case. For this to possible, 19.1 will fall to the right of the mean of the distribution.

Therefore,

X<19.1i.e<19.1

So, is less than 19.1 years.

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